Certain patterns occur often when applying the multiplication principle. As we saw in Example 2, the factors that result from choices are often the same. In this case, we can use exponents to abbreviate the product:

You may see the factors written with exponents instead of factors so it is important to recognize that they are the same.

Another pattern that results from the multiplication principle can be written using factorial notation. Suppose a production line requires six workers to carry out six different jobs. Each worker can only do one job at a time. Once a worker is selected for a job, the other jobs must be carried out by the remaining workers. To find the number of ways we can assign workers to jobs, calculate the product

The number of ways to make each choice drops by one in each factor since each worker can only do one job. In effect, we can’t choose the same worker twice. This is often indicated by saying that we want to assign workers without repetition.

This type of product occurs so often that it is assigned its own symbol.

Factorial Notation

For any positive integer n,

n! = n (n-1) (n-2) … 3 · 2 · 1

The value of 0! is defined to be 1.

When we read an expression with factorial notation, a symbol like n! is read “n factorial”.

Example 3 Use Factorial Notation

Compute the value of each expression involving factorial notation.

a. 6!

SolutionUse the formula above to get

6! = 6 · 5 · 4 · 3 · 2 · 1

b. 9!

SolutionIt is tedious to multiply the factors out for larger numbers. Instead, use a calculator’s factorial command to find the product. On a TI graphing calculator, start by typing 9. Then press . Choosing 4 inserts the factorial symbol ! from the PRB menu. The value is displayed on the screen.

c.

SolutionIt is not practical to multiply all of the factors in the numerator and denominator. In addition, each of the factors in the fraction may not be calculated individually. If we try to do this the calculator will return an overflow error. Instead, write down some of the factor to see if any patterns emerge:

Every factor in the denominator is also in the numerator. These factors may be reduced to give

How do you count choices using the multiplication principle?

A small cellular provider gives its customers 2 choices of phones to use. They may use an iPhone or a phone that uses the Android operating system. In addition, the company offers three different calling plans: Budget plan, Regular plan, and the Deluxe plan. How many different choices of phone and calling plan does a customer have?

To answer this question, we can use a decision tree and list out all of the choices a customer may make.

A decision tree show the different choice a customer makes when choosing a phone and plan. If we move left to right through the tree, we can list out each of the possibilities:

By listing out each of the possibilities, we see that there are six possible phone/plan choices. The decision tree helps us to list out these possibilities. However, if we only need to know how many choices, we can multiply the number of choices for phones and plans.

This strategy is useful for determining the total number of choices even when there are a larger number of choices.

Multiplication Principle

Suppose we wish to know the number of ways to make n choices where there are

d_{1} ways to make choice 1

d_{2} ways to make choice 2

d_{n} ways to make choice n

Then the total number of ways to make all of the choices is

d_{1} · d_{2} · … · d_{n}

Example 1 Multiplication Principle

An online custom bicycle seller wishes to count the total number of different types of bicycles that are available through its website. The seller offers 4 different frame styles, 8 different fender colors, 10 different tire colors, 8 different wheel colors, 6 different pedal colors, and 12 different accessory colors. How many different bicycles can a customer order?

SolutionEach choice the customer must make leads to a different factor in the multiplication principle.

There are 184,320 different bicycles that can be ordered.

Example 2 Multiplication Principle

As the number of cars on the road has increased, so has the number of license plates. The format of the license plate determines how many different license plates there are. For each of the formats below, find the number of different license plates that are available.

a. Three numbers

SolutionWe use the multiplication principle and choose each number. There are ten choices for numbers 0 through 9 giving

b. Three letters followed by three numbers

SolutionIn this type of license plate, we have six choices to make. For each of the first three choices, there are 26 letters to choose from. For the last three choices, there are 10 numbers to choose from. This gives leads to the total number of license plates,

c. Six characters where each character may be a letter or number

SolutionSince the character can be a letter or a number, there are 36 choices for each character. This gives a total number of license plates,

How is Bayes’ Rule used to compute conditional probability?

In Question 2, we learned that the likelihood of an event A occurring given that an event has already occurred is

We can also use the same basic expression to find the likelihood of an event B occurring given that an event A has already occurred,

Each of these expressions may be solved for the joint probability in the numerator to give

The joint event A and B is exactly the same event as the joint event B and A. This means their probabilities are also the same. Setting the left sides of these expressions equal gives

We can solve for either conditional probability, but if we solve for P(B | A) we get the most basic form of Bayes’ Rule.

Bayes’ Rule

If A and B are events, the conditional probability P(B | A) may be computed in terms of using

This expression allows us to compute one conditional probability in terms of the “reverse” conditional probability. In practice, the most challenging part of using Bayes’ Rule is identifying the events and computing the probabilities on the right side. We can simplify this task using a tree diagram.

For instance, let’s return to the tax return tree diagram we developed in Question 4.

This tree diagram is defined in terms of the marginal probabilities P(E) and P(E’), as well as the conditional probabilities P(R | E), P(R’ | E), P(R | E’), and P(R’ | E’). If we want to find the likelihood of one of these conditional probabilities reversed such as P(E | R), we apply Bayes’ Rule to give

Example 6 Bayes’ Rule

The tree diagram for tax returns is shown below.

Using the events

E: return is selected for further examination

E’: return is not selected for further examination

R: return results in a refund of taxes paid

R’: return does not result in a refund of taxes paid

find the probability that a return was examined if we already know it resulted in a refund.

SolutionWe want to find P(E | R). Since the tree diagram is drawn to correspond to events given E or E’, we’ll apply Bayes’ Rule to “reverse” the conditional probabilities. For this conditional probability, Bayes’ Rule gives us

We can locate and highlight these probabilities on the tree diagram.

The probabilities in the numerator are located along the green branch. The probability in the denominator is found using the red and green branches which all terminate at R. Put the numbers in to yield

If a return results in a refund, it is unlikely the return was examined.

The rule for computing conditional property can be interpreted different. In Question 2, we defined the conditional probability . If we multiply both side of this equation by P(B), we get

We can also apply this strategy to the conditional probability to obtain a similar expression,

These expressions give the joint probability of A and B as a product of a conditional probability and a marginal probability.

Product Rule for Probability

The probability of the event A and B is

P( A and B) = P(A | B) P(B)

or

P(A and B) = P(B | A) P(A)

Note that we have used the fact collection of outcomes in A and B is the same as the collection of outcomes in B and A.

This rule is very useful when we work with tree diagrams. Let’s return to the consumer cell phone survey. We constructed the tree diagram for this survey shown below.

In this tree diagram, a branch passing through the event M, “consumer is male”, and the event B, “consumer owns a basic phone”, is colored red. The first part of the branch ending at male is labeled with the probability 0.551. This is the probability that the consumer is male, P(M) = 0.551. The part of the branch beginning with male and ending at basic phone is labeled with the probability 0.316. This is the probability that the consumer owns a basic phone given that the consumer is male, P(B | M) = 0.316. Using the Product Rule for Probability, we may multiply these values to give

In words, the product of the probabilities along the branch gives us the likelihood that a consumer owns a basic phone and is male.

Product Rule for Tree Diagrams

The product of all probabilities along a branch on a tree diagram is the likelihood of all events occurring that are on the branch.

Example 5 Product Rule for Tree Diagrams

In 2008, 1.01% of all individual tax returns were selected for further examination. Of those examined, 3.59% received a refund on taxes paid. Of the returns that were not examined, 86.44% received a refund on taxes paid. Use this information to answer the parts below.

a. Label the branches on the tree diagram below.

SolutionTo make it easier to work with the events, define the events below:

E: return is selected for further examination

E’: return is not selected for further examination

R: return results in a refund of taxes paid

R’: return does not result in a refund of taxes paid

Using complements allows us to limit the number of letters used to represent the different events. Using these events and the fact that 1.01% of returns are selected for further examination, we write P(E) = 0.0101. In addition, we can use the fact that the sum of probabilities from any event must be 1 (P(E) + P(E’) = 1 ). This tells us that P(E’) = 0.9899 . Label these branches on the tree diagram.

Now let’s look at the statement, “Of those examined, 3.59% received a refund on taxes paid.” This tells us that P(R | E) = 0.0359. The other branch originating at E is the complement, so P(R’ | E) = 1 – 0.0359 = 0.9641. A similar strategy gives us the probabilities originating at E’,

This gives the completed tree diagram below.

b. Find the likelihood that a return is not selected for examination and the return does not result in a refund.

SolutionThe event we are interested in is E’ and R’. These individual events are highlighted in the tree diagram below.

The joint probability is the product of the probabilities along the highlighted branch,

c. Find the probability that a return results in no refund.

SolutionTwo branches along the tree terminate at returns that result in no refund.

As we saw in part b, returns along the lower branch are not examined and result in no refund (E’and R’). Returns along the upper branch are examined and result in no refund (E and R’). Together, these branches give all returns that result in no return. In effect, the returns in R’ are the same as the returns in (E’ and R’) or (E and R’). Since there are no returns that correspond to both branches, we can find the likelihood of a return resulting in no refund by applying the product rule to each branch and then add the results:

In general, there are no outcomes in common to branches corresponding joint probabilities in a tree diagram. This means we may add probabilities from these branches to create other events that are compound events created with “or”. This strategy will be useful for computing other marginal probabilities for Bayes’ Rule in Question 5.

When computing conditional probabilities, you might be curious to know whether the fact that an event has occurred has any effect on the probability of another event. For instance, earlier in this section we computed the probability of a consumer owning a smartphone as P(S) ≈ 0.60 . We also computed the likelihood of a consumer owning a smartphone given the fact that the consumer is male, P(S | M) ≈ 0.684. The fact that the consumer is male changes the likelihood of the consumer owning a smartphone. These events are an example of dependent events since one event occurring changes the likelihood of another event occurring. When two events are independent, one event occurring has no effect on the likelihood of the other event occurring.

Independent Events

If one event occurring does not change the likelihood of another event occurring, the two events are independent. This means that for events A and B,

P(A | B) = P(A) and P(B | A) = P(B)

Example 4 Are the Events Independent?

In Example 1, we calculated the conditional probabilities

where F and B represent the events,

F: consumer is female B: consumer owns a basic phone

Are F and B independent events?

SolutionWe already have the conditional events. To determine if P(B | F) = P(B) and P(F | B) = P(F), we need to compute P(B) and P(F). The information for these events are given in the table below.

The likelihood of “consumer is female” is calculated by dividing the number of female consumers by the number of consumers surveyed,

Since , the likelihood of F changes if B has occurred. Therefore, F and B are dependent.

Similarly, the likelihood that the “consumer owns a basic phone” is This is different from the conditional probability .

To prove that two events like B and F are dependent, we only need to show that P(F | B) ≠ P(F) or P(B | F) ≠ P(B). However, to prove that the events are independent, we must prove that both P(F | B) = P(F) and P(B | F) = P(B) .